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# M07 Q24

Author Message
Manager
Joined: 17 May 2010
Posts: 121

Kudos [?]: 42 [0], given: 5

Location: United States
Concentration: Entrepreneurship, Marketing
Schools: USC (Marshall) - Class of 2013
GMAT 1: 770 Q50 V46
GPA: 3.26
WE: Brand Management (Consumer Products)

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24 Sep 2010, 01:59
Quote:
A plane can fly from Europe to North America in 12.4 hours, but the return trip only takes 10.26 hours. If the average ground speed of a plane is 700 km/h, what is the approximate distance between Europe and North America?

A. 8680
B. 8273
C. 7931
D. 7245
E. 7131

I first solved it in the straightforward way -- average time (11.33 hours) and then multiply by 700km/h.

I was confused after solving this, however, as weighted rates skew off the average. For example, if I walk from home to work at 4mph, but then run home from work at 6mph, my average speed isn't 5mph -- it's 4.8mph, because I spent more time walking the 4mph trip.

Could anybody explain why that situation doesn't apply here?
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Last edited by NewSc2 on 24 Sep 2010, 12:46, edited 1 time in total.

Kudos [?]: 42 [0], given: 5

Intern
Joined: 11 Aug 2010
Posts: 23

Kudos [?]: 33 [0], given: 37

Schools: SUNY at Stonyb Brook
WE 1: 4 yrs

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24 Sep 2010, 08:52
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps.
btw is the answer ~ 9064 km approximately?
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Kudos [?]: 33 [0], given: 37

Manager
Joined: 17 May 2010
Posts: 121

Kudos [?]: 42 [0], given: 5

Location: United States
Concentration: Entrepreneurship, Marketing
Schools: USC (Marshall) - Class of 2013
GMAT 1: 770 Q50 V46
GPA: 3.26
WE: Brand Management (Consumer Products)

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24 Sep 2010, 12:46
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps.
btw is the answer ~ 9064 km approximately?

No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.
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Kudos [?]: 42 [0], given: 5

Senior Manager
Joined: 18 Aug 2009
Posts: 414

Kudos [?]: 145 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0

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11 Apr 2011, 23:14
NewSc2 wrote:
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps.
btw is the answer ~ 9064 km approximately?

No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.

Can any of experts comment on this problem. I have problem understanding the concept as well. This problem should be solved using the formula Average Distance= Speed * Time, since the time and speed are different for two trips (from A to b and from b to A because there was the speed of Earth which rotates one way), I am trying to use the formula Average Distance= (700+v)*Time, where v is the speed of the Earth's rotation as plane moves from A to B.
However, the OA solution to this problem is very straightforward, taking the average of time and multiplying it by 700.
Any help with this problem would be appreciated. Thank you.
_________________

Never give up,,,

Kudos [?]: 145 [0], given: 16

Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 871

Kudos [?]: 401 [0], given: 123

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12 Apr 2011, 00:45
I got C.

2d / 700 =total time

Or this is the same as d = average speed * average time. = 700 * 11.33

Kudos [?]: 401 [0], given: 123

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1965

Kudos [?]: 2099 [1], given: 376

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12 Apr 2011, 00:49
1
KUDOS
mirzohidjon wrote:
NewSc2 wrote:
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps.
btw is the answer ~ 9064 km approximately?

No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.

Can any of experts comment on this problem. I have problem understanding the concept as well. This problem should be solved using the formula Average Distance= Speed * Time, since the time and speed are different for two trips (from A to b and from b to A because there was the speed of Earth which rotates one way), I am trying to use the formula Average Distance= (700+v)*Time, where v is the speed of the Earth's rotation as plane moves from A to B.
However, the OA solution to this problem is very straightforward, taking the average of time and multiplying it by 700.
Any help with this problem would be appreciated. Thank you.

$$Speed_{Average} = \frac{Total \hspace{2} Distance}{Total \hspace{2} Time}$$

$$Speed_{Average} = \frac{Total \hspace{2} Distance}{Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}+Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}}$$

$$Speed_{Average} = 700$$

Let the distance between A to B be D.
$$Total \hspace{2} Distance = D+D=2D$$

$$Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}=12.4$$

$$Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}=10.26$$

$$Total \hspace{2} Time = 12.4+10.26=22.66$$

$$700=\frac{2D}{22.66}$$

$$D=\frac{700*22.66}{2}=7931$$

Ans: "C"
****************************************

"Average Distance= (700+v)*Time". I am not aware of any such formula.
Moreover, distance between 2 points A and B is constant. What is average distance then, I wonder!!!

700= Average Speed of the plane both ways.
Earth's speed = $$\nu_{e}$$
Plane's speed = $$\nu_{p}$$

A to B: Speed= Plane's speed-Earth's speed=$$\nu_{p}-\nu_{e}$$
B to A: Speed= Plane's speed+Earth's speed=$$\nu_{p}+\nu_{e}$$

From A to B:
$$Time=\frac{Distance}{Speed}$$
$$12.4=\frac{D}{\nu_{p}-\nu_{e}}$$

From B to A:
$$Time=\frac{Distance}{Speed}$$
$$10.26=\frac{D}{\nu_{p}+\nu_{e}}$$

Since we don't know either the speed of plane or the speed of the earth, proceeding this way would prove futile.
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Kudos [?]: 2099 [1], given: 376

Senior Manager
Joined: 18 Aug 2009
Posts: 414

Kudos [?]: 145 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0

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12 Apr 2011, 01:47
fluke, Thank you for nice explanation. When I meant v, i thought of the speed of Earth. After reading your post, i understood that 700, which is average speed of the plane on the ground already includes that (speed of the Earth). Now, I understand why we use the formula 2D/(t1+t2) to get the average speed.
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Kudos [?]: 145 [0], given: 16

Re: M07 Q24   [#permalink] 12 Apr 2011, 01:47
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# M07 Q24

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